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The Dirac Delta 1 Supplement. The , A Cautionary Tale

Note. In the study of charge distributions in electricity and magnetism, when con- sidering point charges it is common to introduce the “Dirac delta function,” δ(x). This function is defined to be extended real valued with the following properties:   0 if x 6= 0 Z ∞ δ(x) = and δ(x) dx = 1.  ∞ if x = 0 −∞

However, this is very misleading! First, Riemann do not address extended real valued functions. Second, such a function would violate Proposition 4.9 of Royden and Fitzpatrick’s , 4th Edition: Proposition 4.9. Let f be a nonnegative measurable (extended real valued) function on E. Then the Lebesgue E f = 0 if and only if f = 0 a.e. on E.

Note. In this supplement, we refer to the junior-level undergraduate text Foun- dations of Electromagnetic Theory, 3rd Edition by John Reitz, Frederick Milford, Robert Christy, Addison-Wesley (1979). We reveal some errors in it’s presentation of δ(x) and explore an alternative approach to “get the math right.” The Dirac Delta Function 2

Note. Reitz, Milford, and Christy introduce the Dirac delta function on pages 43 and 44.

Note. The motivation for δ(x) is given on page 43. On page 44, they comment that “. . . it is a legitimate mathematical object, which leads to no difficulties if one is cautious. . . .” In the footnote, they falsely state: “The of such a function is zero if it exists at all, but the integration can be handled by the more general Lebesgue integral.” Well, the Lebesgue integral does “handle” the integration, but it gives zero as the integral value. The Dirac Delta Function 3

Definition. Reitz, Milford, and Christy give more details in their Appendix IV, and the notation there is consistent with that used above in these notes (as opposed to a function of a vector as stated on their pages 43 and 44).

Note. In equations IV-2, IV-3, and IV-4 they write δ(x) as the of three

2 2 √1 −x /ε different functions. In equation IV-2, they state δ(x) = limε→0 πεe (a normal √ distribution with µ = 0 and σ = ε/ 2). Since this is a

∞ 2 2 R √1 −x /ε , −∞ πεe dx = 1. As ε → 0, the distribution gets very narrow with a large spike at x = 0. If we replace ε with 1/n where n ∈ N, then we 2 √n −(nx) can create a of functions fn = π e where limn→∞ fn(x) = δ(x). The Dirac Delta Function 4

Note. Recall that from Royden and Fitzpatrick:

Monotone Convergence Theorem. Let {fn} be an increasing

sequence of nonnegative measurable functions on set E. If {fn} → f pointwise a.e. on E, then

Z  Z   Z lim fn = lim fn = f. n→∞ E E n→∞ E

2 √n −(nx) In the previous note, we have fn = π e and limn→∞ fn(x) = δ(x), but the convergence is not monotone increasing, so the Monotone Convergence Theorem R ∞ does not apply to give that the integral −∞ δ(x) dx is 1. It is clear that Reitz, Milford, and Christy are trying to use the values of the integrals of the normal distributions to justify the claim that the integral of δ(x) should be 1, but this is not the case.

Note. Recall that from Royden and Fitzpatrick:

Fatou’s Lemma. Let {fn} be a sequence of nonnegative measurable

functions on set E. If {fn} → f pointwise a.e. on E, then Z Z  f ≤ lim inf fn . E E Now we can apply Fatou’s Lemma to the sequence of normal distributions above, since the convergence to δ(x) is pointwise and all functions here are nonnegative. But this simply allows us to conclude that

∞ ∞ Z Z n 2 δ(x) dx ≤ √ e−(nx) dx = 1, −∞ −∞ π still not an equality of 1, but merely an upper bound of 1. The Dirac Delta Function 5

Note. Reitz, Milford, and Christy’s “no difficulties if one is cautious” comment is correct, with the right kind of caution! In Analysis 2 (MATH 4227/5227) Riemann- Stieltjes integrals are introduced (see my online class notes on 6-3. The Riemann- Stieltjes Integral; in particular, see Theorem 6-26). One way to resolve the desired properties of the Dirac delta function, is to use Riemann-Stieltjes integrals. You should review these notes before reading the next note.

Note. Let   0 for x ∈ (−∞, 0) g(x) =  1 for x ∈ [0, ∞). Then the of g is 0 if n 6= 0. The definition of limit gives that the derivative of g is ∞ at 0. Also, we have the Riemann-Stieltjes integral Z ∞ Z 0   Z ∞ dg = dg + (1) lim g(x) − lim g(x) + dg = 0 + (1)[1] + 0 = 1. −∞ −∞ x↓0 x↑0 0 So the derivative of g(x) has the values of δ(x) given above and the Riemann- Stieltjes integral of f(x) = 1 with respect to g satisfies the integral property of δ(x) given above. If f is a function defined on all of R, then we can use the Riemann-

Stieltjes integral to determine the value of f at a specific point (say x = x0): Z ∞   f(x) dg(x − x0) = f(x0) lim g(x − x0) − lim g(x − x0) = f(x0)[1] = f(x0). −∞ x↓x0 x↑x0

Note. Another, more sophisticated solution involves the concen-

trated at x0, denoted δx0 . For a σ- M of of a set X and a point x0 belonging to X, the measure of 1 is assigned to a set in M that contains x0, and a measure of 0 is assigned to a set that does not contain x0. We then have for R f on X that X f dδx0 = f(x0), as desired. This is explored in Problem 18.26 (in 18.2. Integration of Nonnegative Measurable Functions) and Problem 18.27(ii) (in 18.3. Integration of General Measurable Functions). The Dirac Delta Function 6

Note. The Dirac delta function is named for Paul A. M. Dirac (August 8, 1902– October 20, 1984). Another approach is to treat δ(x) not as a function, but as a distribution (or a ). It is therefore more accurately called the “Dirac delta distribution.” A classical work on the theory of distributions is I. M. Gel’fand and G. E. Shilov’s Generalized Functions, 15, Academic Press, (1966-1968). Revised: 11/9/2020